求通项公式

aimisiyou

求解通项公式。

王守恩

再来几项？
a(2)=0, a(3)=?, a(4)=?, a(5)=?,......

aimisiyou

依次递归都可以推出啊，如a(3)=1/2,    a(4)=5/2,...

northwolves

方括号是取整？

northwolves

$a_{i+1}=\frac{i(i+1)+2i-10}{4}+\frac{2}{2^i}+\frac{1}{2}a_{i}$

aimisiyou

northwolves 发表于 2021-5-15 11:18
$a_{i+1}=\frac{i(i+1)+2i-10}{4}+\frac{2}{2^i}+\frac{1}{2}a_{i}$

是这个递推关系。

northwolves

$a_{i+1}=\frac{i^2+3i-10}{4}+\frac{1}{2^{i-1}}+\frac{1}{2}a_{i}$

$=\frac{-i(i+1)+2i*(i-1)-6((i+1)-2i)+4(1-2)}{4}-\frac{1-2}{2^{i-1}}+\frac{1}{2}a_{i}$

$a_{i+1}+\frac{i(i+1)+6(i+1)-4}{4}+\frac{1}{2^{i-1}}=\frac{1}{2}(a_{i}+\frac{i(i-1)+6i-4}{4}+\frac{1}{2^{i-2}})$

令$b_{i}=a_{i}+\frac{i(i-1)+6i-4}{4}+\frac{1}{2^{i-2}}$

则$b_{2}=0+\frac{2*(2-1)+6*2-4}{4}+\frac{1}{2^{2-2}}=\frac{7}{2}$,$b_{n}=\frac{1}{2}b_{n-1}=...=\frac{7}{2^{n-1}$

代入 $a_{i}+\frac{i(i-1)+6i-4}{4}+\frac{1}{2^{i-2}}=\frac{7}{2^{i-1}$

$a_{n}=1+\frac{5}{2^{n-1}}-\frac{n(n+5)}{4}$

aimisiyou

northwolves 发表于 2021-5-15 14:05
$a_{i+1}=\frac{i^2+3i-10}{4}+\frac{1}{2^{i-1}}+\frac{1}{2}a_{i}$

$=\frac{-i(i+1)+2i*(i-1)-6((i+1) ...

不对啊   你这通项公式a(3)不等于1/2啊？

王守恩

aimisiyou 发表于 2021-5-15 10:39
依次递归都可以推出啊，如a(3)=1/2,    a(4)=5/2,...

1，RecurrenceTable[{a[2] = 0, a[n + 1] = (a[n] - 3 + n + 4 (1/2)^n - 2 + n (n + 1)/2)/2}, a, {n, 2, 10}]
{0, 1/2, 5/2, 47/8, 21/2, 521/32, 741/32, 3979/128, 2563/64}

2，RecurrenceTable[{a[2] = 0, a[n + 1] = Floor[(a[n] - 3 + n + 4 (1/2)^n - 2 + n (n + 1)/2)/2]}, a, {n, 2, 95}]
{0, 0, 2, 5, 10, 16, 23, 31, 40, 50, 61, 73, 86, 100, 115, 131, 148, 166, 185, 205,
226, 248, 271, 295, 320, 346, 373, 401, 430, 460, 491, 523, 556, 590, 625, 661, 698,
736, 775, 815, 856, 898, 941, 985, 1030, 1076, 1123, 1171, 1220, 1270, 1321, 1373,
1426, 1480, 1535, 1591, 1648, 1706, 1765, 1825, 1886, 1948, 2011, 2075, 2140, 2206,
2273, 2341, 2410, 2480, 2551, 2623, 2696, 2770, 2845, 2921, 2998, 3076, 3155, 3235,
3316, 3398, 3481, 3565, 3650, 3736, 3823, 3911, 4000, 4090, 4181, 4273, 4366, 4460,}

3，Table[((n + 2) (n - 3) - 4)/2, {n, 2, 95}]
{-4, -2, 1, 5, 10, 16, 23, 31, 40, 50, 61, 73, 86, 100, 115, 131, 148, 166, 185, 205,
226, 248, 271, 295, 320, 346, 373, 401, 430, 460, 491, 523, 556, 590, 625, 661, 698,
736, 775, 815, 856, 898, 941, 985, 1030, 1076, 1123, 1171, 1220, 1270, 1321, 1373,
1426, 1480, 1535, 1591, 1648, 1706, 1765, 1825, 1886, 1948, 2011, 2075, 2140, 2206,
2273, 2341, 2410, 2480, 2551, 2623, 2696, 2770, 2845, 2921, 2998, 3076, 3155, 3235,
3316, 3398, 3481, 3565, 3650, 3736, 3823, 3911, 4000, 4090, 4181, 4273, 4366, 4460,}

4，LinearRecurrence[{3, -3, 1}, {1, 5, 10}, 92]
{1, 5, 10, 16, 23, 31, 40, 50, 61, 73, 86, 100, 115, 131, 148, 166, 185, 205,
226, 248, 271, 295, 320, 346, 373, 401, 430, 460, 491, 523, 556, 590, 625, 661, 698,
736, 775, 815, 856, 898, 941, 985, 1030, 1076, 1123, 1171, 1220, 1270, 1321, 1373,
1426, 1480, 1535, 1591, 1648, 1706, 1765, 1825, 1886, 1948, 2011, 2075, 2140, 2206,
2273, 2341, 2410, 2480, 2551, 2623, 2696, 2770, 2845, 2921, 2998, 3076, 3155, 3235,
3316, 3398, 3481, 3565, 3650, 3736, 3823, 3911, 4000, 4090, 4181, 4273, 4366, 4460}

5，RecurrenceTable[{a[1] = 1, a[2] = 5, a[3] = 10, a[n + 1] = 3 a[n] - 3 a[n - 1] + a[n - 2]}, a, {n, 1, 92}]
{1, 5, 10, 16, 23, 31, 40, 50, 61, 73, 86, 100, 115, 131, 148, 166, 185, 205,
226, 248, 271, 295, 320, 346, 373, 401, 430, 460, 491, 523, 556, 590, 625, 661, 698,
736, 775, 815, 856, 898, 941, 985, 1030, 1076, 1123, 1171, 1220, 1270, 1321, 1373,
1426, 1480, 1535, 1591, 1648, 1706, 1765, 1825, 1886, 1948, 2011, 2075, 2140, 2206,
2273, 2341, 2410, 2480, 2551, 2623, 2696, 2770, 2845, 2921, 2998, 3076, 3155, 3235,
3316, 3398, 3481, 3565, 3650, 3736, 3823, 3911, 4000, 4090, 4181, 4273, 4366, 4460}

6，Table[(n + 3)/2^(n - 1), {n, 2, 20}]   说明：6=1-2   n=3,4,5,6,7,8,9,...
{5/2, 3/2, 7/8, 1/2, 9/32, 5/32, 11/128, 3/64, 13/512, 7/512, 15/2048, 1/256, 17/8192,
9/8192, 19/32768, 5/16384, 21/131072, 11/131072, 23/524288}

aimisiyou

northwolves 发表于 2021-5-15 14:05
$a_{i+1}=\frac{i^2+3i-10}{4}+\frac{1}{2^{i-1}}+\frac{1}{2}a_{i}$

$=\frac{-i(i+1)+2i*(i-1)-6((i+1) ...

为什么你的方法看起来没问题，但结果却不对呢？